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Illustrative examples:
This equation of the form 𝑦 ′ + 𝑝(𝑥)𝑦 = 𝑞(𝑥) is called a linear differential equation, where 𝑝(𝑥) and 𝑞(𝑥) are continuous functions on some interval. 90 www. kokanduni.uz According to Bernoulli's method, we look for the solution of equation (1) in the form of the product of two unknown functions 𝑢 = 𝑢(𝑥) and 𝑣 = 𝑣(𝑥) : 𝑦 = 𝑢𝑣(𝑦(𝑥) = 𝑢(𝑥) ⋅ 𝑣(𝑥)) (1) If we substitute 𝑦 = 𝑢𝑣, 𝑦 ′ = (𝑢𝑣) ′ = 𝑢 ′ 𝑣 + 𝑢𝑣 ′ above for 𝑦 and 𝑦 ′ in equation (1), then its derivative will be 𝑋 1 ′ = (𝑢𝑣) ′ = 𝑢 ′ 𝑣 + 𝑢𝑣 ′ . It seems absurd to introduce two unknown functions 𝑢 = 𝑢(𝑥), 𝑣 = 𝑣(𝑥) instead of the unknown function 𝑦 = 𝑦(𝑥) to be found. However, we will see later that the appropriate selection of one of the functions 𝑢 = 𝑢(𝑥) and 𝑣 = 𝑣(𝑥) allows us to easily find the other. As a result, 𝑦(𝑥) is found. After substitution, the differential equation 𝑢 ′ 𝑣 + 𝑢𝑣 ′ + 𝑝𝑢𝑣 = 𝑞 is formed. Now we choose the function 𝑣 ′ + 𝑝𝑣 = 0 in such a way (using the option of choosing this function arbitrarily) that we have 𝑢 ′ 𝑣 + 𝑢 ( 𝑣 ′ + 𝑝𝑣 ) = 𝑞 . This differential equation comes to this variable A=C separable equation. Integrating the following equation 𝑑𝑣 𝑣 = −𝑝𝑑𝑥 we find: ∫ 𝑑𝑣 𝑣 = −∫𝑝𝑑𝑥, 𝑙𝑛 𝑣 = −∫𝑝𝑑𝑥 𝑣 = 𝑒 −∫𝑝𝑑𝑥 (2) As a result, the differential equation (2) this takes the following view 𝑢 ′ 𝑒 −∫𝑝𝑑𝑥 = 𝑞 We will solve it: 𝑒 −∫𝑝𝑑𝑥 ⋅ 𝑑𝑢 𝑑𝑥 = 𝑞, 𝑑𝑢 𝑑𝑥 = 𝑞𝑒 ∫𝑝𝑑𝑥 , 𝑑𝑢 = 𝑞𝑒 ∫𝑝𝑑𝑥 𝑑𝑥, 𝑢 = ∫𝑞𝑒 ∫𝑝𝑑𝑥 𝑑𝑥 + 𝐶. (3) From relations (3) and (4) it follows that 𝑦 = 𝑢𝑣 = (∫ 𝑞𝑒 ∫ 𝑝𝑑𝑥 𝑑𝑥 + 𝐶)𝑒 −∫ 𝑝𝑑𝑥 . (4) So, 𝑦 ′ + 𝑝𝑦 = 𝑞 is a generalization of the linear differential equation the solution is 𝑦 = 𝑢𝑣 = (∫ 𝑞𝑒 ∫ 𝑝𝑑𝑥 𝑑𝑥 + 𝐶)𝑒 −∫ 𝑝𝑑𝑥 . (5) Example 1, find the general solution of this equation 𝑦 ′ + 𝑥𝑦 = 𝑥 3 . We find the general solution of this equation using formula (5). 𝑦 = (∫ 𝑥 3 𝑒 ∫ 𝑥𝑑𝑥 𝑑𝑥 + 𝐶)𝑒 −∫ 𝑥𝑑𝑥 = (∫ 𝑥 3 𝑒 𝑥 2 2 𝑑𝑥 + 𝐶) 𝑒 − 𝑥 2 2 = 𝑥 2 − 2 + 𝐶𝑒 − 𝑥 2 2 Therefore, the general solution of the given equation is 𝑦 = 𝑥 2 − 2 + 𝐶𝑒 − 𝑥2 2 . Yüklə 15,42 Mb. Dostları ilə paylaş:
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